\(\overrightarrow{IA}=\left(2;0\right);\overrightarrow{IB}=\left(-2;3\right);\overrightarrow{IC}=\left(x_C-1;y_C+1\right)\)
Ta có : \(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\) (\(I\) là trọng tâm \(\Delta ABC\))
\(\Leftrightarrow\overrightarrow{IC}=-\overrightarrow{IA}-\overrightarrow{IB}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_C-1=-2+2\\y_C+1=0-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_C=1\\y_C=-4\end{matrix}\right.\) \(\Rightarrow C\left(1;-4\right)\)
\(\overrightarrow{AB}=\overrightarrow{DC}\left(hbh\right)\)
\(\Leftrightarrow\left(-4;3\right)=\left(1-x_D;-4-y_D\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_D=5\\y_D=-7\end{matrix}\right.\) \(\Rightarrow D\left(5;-7\right)\)
\(\overrightarrow{OA}+\overrightarrow{OC}=\overrightarrow{0}\) (O là trung điểm AC)
\(\Leftrightarrow\left(3-x_O;-1-y_O\right)=-\left(1-x_O;-4-y_O\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-x_O=x_O-1\\-1-y_O=y_O+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_O=2\\y_O=-\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow O\left(2;-\dfrac{5}{2}\right)\)